Leo Giugiuc's Cyclic Inequality with a Constraint
Problem
Solution 1
Homogenize the inequality as
$\displaystyle \begin{align} &\frac{a^3+b^3+c^3}{abc}+\frac{16abc}{(a+b)(b_c)(c+a)}\ge 5&\Leftrightarrow\\ &\frac{a^3+b^3+c^3+3abc}{abc}+\frac{16abc}{(a+b)(b+c)(c+a)}\ge 8&\Leftrightarrow\\ &\frac{a^3+b^3+c^3+3abc}{2abc}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 4&\Leftrightarrow \end{align}$
Set $\displaystyle \frac{2a}{b+c}=x\gt 0,$ $\displaystyle \frac{2b}{c+a}=y\gt 0,$ $\displaystyle \frac{2c}{a+b}=z\gt 0.$ Then $\displaystyle xy+yx+zx+\frac{8abc}{(a+b)(b+c)(c+a)}=4.$ So the latter inequality is equivalent to
$\displaystyle \frac{a^3+b^3+c^3+3abc}{2abc}\ge xy+yz+zx.$
Further,
$\displaystyle\begin{align} &\frac{a^3+b^3+c^3+3abc}{2abc}\ge xy+yz+zx&\Leftrightarrow\\ &\frac{a^3+b^3+c^3+3abc}{2abc}\ge\frac{4}{(a+b)(b+c)(c+a)}\cdot [ab(a+b)+bc(b+c)+ca(c+a)]. \end{align}$
By the AM-GM inequality, $\displaystyle \frac{1}{2abc}\ge\frac{4}{(a+b)(b+c)(c+a)},$ implying
$\displaystyle\begin{align}&\frac{1}{2abc}\cdot [ab(a+b)+bc(b+c)+ca(c+a)]\\ &\qquad\ge\frac{4}{(a+b)(b+c)(c+a)}\cdot [ab(a+b)+bc(b+c)+ca(c+a)].\end{align}$
Thus, suffice it to show that
$\displaystyle \frac{a^3+b^3+c^3+3abc}{2abc}\ge \frac{1}{2abc}\cdot [ab(a+b)+bc(b+c)+ca(c+a)],$
which is equivalent to
$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a),$
which is Schur's inequality of degree 3.
Solution 2
$\displaystyle \begin{align} &a^3+b^3+c^3\geq \frac{(a+b+c)^3}{9}~\text{(Jensen's)} \\ &\frac{16}{(a+b)(b+c)(c+a)}\geq \frac{54}{(a+b+c)^3}~\text{(AM-GM)}. \end{align}$
$\displaystyle x~\overset{\text{def}}{=} ~\frac{(a+b+c)^3}{9} \geq 3abc = 3~\text{(AM-GM)}.$
From the above three inequalities,
$\displaystyle\begin{align} a^3+b^3+c^3+\frac{16}{(a+b)(b+c)(c+a)}-5&\geq x+\frac{6}{x}-5\\ &=\frac{(x-2)(x-3)}{x}\geq 0. \end{align}$
Acknowledgment
Leo Giugiuc has kindly emailed me a problem of his, along with his own solution.
Solution 2 is by Amit Itagi.
Inequalities with the Product of Variables as a Constraint
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- A Cyclic But Not Symmetric Inequality in Four Variables $\left(\displaystyle 5(a+b+c+d)+\frac{26}{abc+bcd+cda+dab}\ge 26.5\right)$
- Problem 2, the 36th IMO (1995) $\left(\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac{3}{2}\right)$
- An Inequality with Two Cyclic Sums $\left(\displaystyle \sum_{cycl}(a+\sqrt[3]{a}+\sqrt[3]{a^2})\ge 9\sum_{cycl}\frac{1}{1+\sqrt[3]{b^2}+\sqrt[3]{c}}\right)$
- The Roads We Take $(x(x-3(y+z))^2+(3x-(y+z))^2(y+z)\ge 27)$
- Long Huynh Huu's Inequality and Solution $\left(\displaystyle \sum_{i=1}^n\arctan x_i\le (n-1)\frac{\pi}{2}\right)$
- An Inequality with Integrals $\left(\Omega=\int_0^u\frac{u^4+7x^3-25x^2+37x+4}{x^4-3x^3+5x^2-3x+4}dx,\;\Omega(a)+\Omega(b)+\Omega(C)\ge3+5\ln\prod_{cycl}(1+a^2)\right)$
- An Inequality with Cycling Sums $\left(\displaystyle\sum_{cycl}(x^4+y^3+z)\ge \sum_{cycl}\frac{x^2+y^2}{z}+3\right)$
- Two Products: Constraint and Inequality $\left(\displaystyle\prod_{k=1}^n(1+a_k)\ge 2^n\right)$
- Leo Giugiuc's Cyclic Inequality with a Constraint $\left(\displaystyle a^3+b^3+c^3+\frac{16}{(a+b)(b+c)(c+a)}\ge 5\right)$
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